ME242 – MECHANICAL ENGINEERING SYSTEMS
LECTURE 10:
• Laplace transform
7.2
191
ME242 - Spring 2006 - Eugenio Schuster
Time domain (t domain)
LAPLACE TRANSFORM-SOLVING ODEs
Linear
Systems
Differential
equation
Complex frequency domain
(s domain)
Laplace
transform L
Classical
techniques
Response
signal
Algebraic
equation
Algebraic
techniques
Inverse Laplace
transform L-1
Response
transform
The diagram commutes
Same answer whichever way you go
ME242 - Spring 2006 - Eugenio Schuster
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1
LAPLACE TRANSFORM
Exercise: Find the Laplace transform V(s)
dv(t )
+ 6v(t ) = 4u (t )
dt
v(0−) = −3
V (s) =
4
3
−
s (s + 6 ) s + 6
Exercise: Find the Laplace transform V(s)
d 2v(t )
dv(t )
+4
+ 3v(t ) = 5e − 2t
2
dt
dt
v (0−) = −2, v' (0−) = 2
V ( s) =
5
2
−
(s + 1)(s + 2)(s + 3) s + 1
What about v(t)?
193
Time domain (t domain)
ME242 - Spring 2006 - Eugenio Schuster
Linear
cct
Differential
equation
Laplace transforms
Complex frequency domain
(s domain)
Laplace
transform L
Classical
techniques
Response
signal
Algebraic
equation
Algebraic
techniques
Inverse Laplace
transform L-1
Response
transform
The diagram commutes
Same answer whichever way you go
ME242 - Spring 2006 - Eugenio Schuster
194
2
INVERTING LAPLACE TRANSFORM IN PRACTICE
We have a table of inverse LTs
Write F(s) as a partial fraction expansion
b sm + bm−1sm−1 + L + b1s + b0
F(s) = m n
an s + an−1sn−1 + L + a1s + a0
(s − z1 )(s − z2 )L(s − zm )
=K
(s − p1 )(s − p2 )L(s − pn )
=
α1
(s − p1 )
α
α
α
α
αq
2
31 +
32
33
+
+
+ ...+
(s − p2 ) (s − p3 ) (s − p3 )2 (s − p3 )3
s − pq
+
(
)
Now appeal to linearity to invert via the table
Surprise!
Nastiness: computing the partial fraction expansion is best
done by calculating the residues
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ME242 - Spring 2006 - Eugenio Schuster
RATIONAL FUNCTIONS
We shall mostly be dealing with Laplace Transforms
which are rational functions – ratios of polynomials in s
F ( s) =
bm s m + bm −1s m −1 + L + b1s + b0
an s n + an −1s n −1 + L + a1s + a0
=K
( s − z1 )( s − z2 )L( s − zm )
( s − p1 )( s − p2 )L ( s − pn )
pi are the poles and zi are the zeros of the function
K is the scale factor or (sometimes) gain
A proper rational function has n≥m
A strictly proper rational function has n>m
An improper rational function has n<m
ME242 - Spring 2006 - Eugenio Schuster
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3
RESIDUES AT SIMPLE POLES
Functions of a complex variable with isolated, finite
order poles have residues at the poles
F ( s) = K
k1
k2
kn
( s − z1 )( s − z2 )L( s − zm )
=
+
+L+
( s − p1 )( s − p2 )L( s − pn ) ( s − p1 ) ( s − p2 )
( s − pn )
(s − pi )F ( s) = k1 ( s − pi ) + k2 ( s − pi ) + L + ki + L + kn ( s − pi )
( s − p1 )
Residue at a simple pole:
( s − p2 )
( s − pn )
ki = lim ( s − pi ) F ( s )
s → pi
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ME242 - Spring 2006 - Eugenio Schuster
RESIDUES AT SIMPLE POLES
Exercise: Find the Laplace transform V(s)
4
3
−
s (s + 6 ) s + 6
4 − 3s
V (s) =
s (s + 6 )
11
2
v(t ) = u (t ) − e −6t u (t )
3
3
dv(t )
+ 6v(t ) = 4u (t )
dt
v(0−) = −3
V (s) =
Exercise: Find the Laplace transform V(s)
d 2v(t )
dv(t )
+4
+ 3v(t ) = 5e − 2t
2
dt
dt
v (0−) = −2, v' (0−) = 2
ME242 - Spring 2006 - Eugenio Schuster
5
2
−
(s + 1)(s + 2)(s + 3) s + 1
5 − 2(s + 2 )(s + 3)
V ( s) =
(s + 1)(s + 2)(s + 3)
1 −t
5
v(t ) = e u (t ) − 5e − 2t u (t ) + e −3t u (t )
2
2
V ( s) =
What about v(t)?
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4
RESIDUES AT MULTIPLE POLES
Let us assume that V(s) has a pole of order n at s=a
Residues at a multiple pole:
km =
1
d m −1 
lim
( s − a ) nV ( s ) ,


−
m
1

( m − 1)! s → a ds
m = 1L n
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ME242 - Spring 2006 - Eugenio Schuster
RESIDUES AT MULTIPLE POLES
2
Example: V ( s ) = 2 s + 5s = k1 + k2 + k3
(s + 1)3 (s + 1)3 (s + 1)2 s + 1
k1 = lim
s → −1
(s + 1)3 2s 2 + 5s 

(s + 1)3
 = −3
 (s + 1)3  2 s 2 + 5s  
d 

 = 1
k2 = lim 

s → −1 ds
3
(
)
+
s
1


 (s + 1)3  2 s 2 + 5s  
d2 
1

 = 2
k3 = lim 2 

3
2! s →−1 ds
+
1
s
(
)


 2 s 2 + 5s  −1 
3
1
2  −t
 = L  −
 = e (− 3t 2 + t + 2 )u (t )
+
+
L−1 
3 
3
2
+
s
1
(
)
(
)
(
)
+
+
+
s
1
s
1
s
1




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RESIDUES AT COMPLEX POLES
Compute residues at the poles
lim ( s − a) F ( s)
s →a
Bundle complex conjugate pole pairs into secondorder terms if you want. But you will need to be
careful!
[
(
( s − α − jβ )( s − α + jβ ) = s 2 − 2αs + α 2 + β 2
)]
Inverse Laplace Transform is a sum of complex
exponentials
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ME242 - Spring 2006 - Eugenio Schuster
RESIDUES AT COMPLEX POLES
Example: V ( s ) =
20(s + 3)
k1
k2
k2*
=
+
+
(s + 1)(s 2 + 2s + 5) s + 1 s + 1 − j 2 s + 1 + j 2
20( s + 3)
k = lim ( s + 1) F ( s ) =
= 10
1 s → −1
s 2 + 2 s + 5 s = −1
5
j π
20( s + 3)
k =
lim
( s + 1 − 2 j ) F ( s) =
= −5 − 5 j = 5 2e 4
2 s → −1 + 2 j
( s + 1)( s + 1 + 2 j ) s = −1 + 2 j
5
5
j π
−j π 

4
20(s + 3)
5 2e
5 2e 4 

 −1  10
L−1 
+
+

=L 
2
 (s + 1)(s + 2 s + 5)
s + 1 s + 1− j2 s + 1 + j2 


5
5
( −1+ j 2 ) t + j π
( −1− j 2 ) t − j π 

4
4
= 10e −t + 5 2e
+ 5 2e
u (t )


5π 

= 10e −t + 10 2e −t cos(2t + )u (t )
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ME242 - Spring 2006 - Eugenio Schuster
4 

6
NOT STRICTLY PROPER LAPLACE TRANSFORMS
Find the inverse LT of V ( s ) =
s 3 + 6s 2 + 12s + 8
s 2 + 4s + 3
Convert to polynomial plus strictly proper rational function
Use polynomial division
s+2
s + 4s + 3
0 .5 0 .5
= s+2+
+
s +1 s + 3
V (s) = s + 2 +
Invert as normal
2

 dδ (t )
v(t ) = 
+ 2δ (t ) + 0.5e −t + 0.5e −3t u (t )
 dt

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